Enthalpy of Neutralization
Calorimetric Study of $HCl$ and $NaOH$
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Lab Manual: Theory & Calculations
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Aim
To determine the enthalpy change ($\Delta H$) of neutralization between a strong acid ($HCl$) and a strong base ($NaOH$).
Principle
Neutralization involves H⁺ and OH⁻ combining to form water. For a strong acid–strong base pair the enthalpy per mole of water formed is essentially constant. Predict its sign and rough magnitude yourself, then measure it from the temperature change.
$q = m \cdot c \cdot \Delta T$
$m = 100\text{g}, c = 4.18 \text{ J/g}\cdot\text{K}$
Procedure
- 1. Measure $50\text{ mL}$ of $1.0M\ HCl$ and $50\text{ mL}$ of $1.0M\ NaOH$.
- 2. Record the Initial Temperature ($T_i$).
- 3. Mix both solutions in the Styrofoam calorimeter.
- 4. Stir gently and record the Maximum Temperature ($T_f$).
🔮 Predict before you experiment
You will mix 50 mL of 1.0 M HCl with 50 mL of 1.0 M NaOH in an insulated calorimeter. Predict the outcome, then lock it to unlock Mix & React.
⚠️ Safety
- Both 1.0 M HCl and 1.0 M NaOH are corrosive — they attack skin and eyes. Wear goggles and gloves; rinse splashes with plenty of water.
- Neutralisation releases heat — the mixture warms up; handle the calorimeter carefully and do not seal it tightly.
- Add base to acid slowly and stir; never point the mouth of a container at your face.
- The product is dilute salt water, but rinse the calorimeter and neutralise waste before disposal.
- Read the thermometer without removing it from the liquid; it is fragile.
Station 1: Preparation
Station 2: Calorimetry
🧮 Experimental Calculations
1. Mass of Mixture ($m$)
$m = V_{acid} + V_{base} = 50\text{mL} + 50\text{mL} = 100\text{mL}$
Density $\approx 1\text{g/mL} \implies$ $m = 100\text{g}$
2. Heat Released ($q$)
$q = m \cdot c \cdot \Delta T$
3. Moles of Water ($n$)
$n = \text{Molarity} \times V(L)$
$n = 1.0\text{M} \times 0.050\text{L} = 0.050\text{ mol}$
4. Enthalpy Change ($\Delta H$)
$\Delta H = -q/n$
Observation Log
Inferred $\Delta H$
Calculated using $q/n$, where $n = 0.05\text{ moles}$ for $HCl + NaOH$.
Reasoning Journal
Your prediction, observations and interpretation will appear here.